Question

# Use the Laplace transform to solve the given system of differential equations. frac{(d^2x)}{(dt^2)}+frac{(d^2y)}{(dt^2)}=frac{t}{2} frac{(d^2x)}{(dt^2)}-frac{(d^2y)}{(dt^2)}=4t x(0) = 5, x'(0) = 0, y(0) = 0, y'(0) = 0

Laplace transform
Use the Laplace transform to solve the given system of differential equations.
$$\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2}$$
$$\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t$$
$$x(0) = 5, x'(0) = 0,$$
$$y(0) = 0, y'(0) = 0$$

2021-01-26

Step 1
Given that
$$\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2} \dots(A)$$
$$\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t \dots(B)$$
$$x(0) = 5, x'(0) = 0,$$
$$y(0) = 0, y'(0) = 0$$
$$\Rightarrow 2\frac{(d^2x)}{(dt^2)}=(t^2+4t)$$
$$\frac{(d^2x)}{(dt^2)}=\frac{(t^2+4t)}{2}$$
$$\frac{(d^2x)}{(dt^2)}=\frac{(t^2)}{2}+2t$$
Step 2
taking Laplace transform on both sides,
$$s2X(s)-sx(0)-x'(0)=\frac{1}{2}+ \frac{2}{s^2} + \frac{2}{s^2}$$
$$s2X(s)-5s=\frac{1}{(s^3)}+\frac{2}{s^2}$$
$$s2X(s)=\frac{(1+2s+5s^4)}{s^5}$$
$$X(s)=\frac{1}{s^5}+\frac{2}{s^4}+\frac{5}{s}$$
taking Inverse Laplace transform
$$x(t)=\frac{t^4}{4!}+\frac{(2t^3)}{3!}+5u(t)$$
Step 3
$$u(t)=1, t\geq0$$
$$=0 , t<0$$
$$\therefore x(t)=\frac{(t^4)}{24} + \frac{(t^3)}{3} + 5 \text{ for } t\geq0$$
And Subtracting equation (B) from (A)
$$\Rightarrow 2\frac{(d^2y)}{(dt^2)}=t^2-4t$$
$$\frac{(d^2y)}{(dt^2)}=\frac{(t^2-4t)}{2}=\frac{(t^2)}{2}-2t$$
Taking ILT
$$s^2Y(s)-sy(0)-y'(0)=\frac{1}{2} \frac{2}{s^3}-\frac{2}{s^2}$$
$$s^2Y(s)=\frac{1}{(s^3)}-\frac{2}{(s^2)}=\frac{(1-2s)}{s^3}$$
$$Y(s)=\frac{(1-2s)}{s^5}$$
Step 4
$$Y(s)=\frac{1}{s^5}-\frac{2s}{(s^5)}$$
$$Y(s)=\frac{1}{s^5}-\frac{2}{(s^4)}$$
Taking Inverse Laplace Transform
$$y(t)=\frac{(t^4)}{4!}-\frac{(2t^3)}{3!}$$
$$=\frac{(t^4)}{24}-\frac{(t^3)}{3}$$
$$y(t)=\frac{(t^4)}{24}-\frac{(t^3)}{3}$$